The marble rolls down the track shown in FIGURE P12.75 and around a loop-the-loop of radius R. The marble has mass m and radius r. What minimum height h must the track have for the marble to make it around the loop-the-loop without falling off?

Radius of track = R

Radius of ball =r

mass of ball =m

Lets draw the diagram as below

Potential energy of the ball at height h is (note ball has radius r)

PE = mg(h+r) .............................(1)

At the top point on the circular track

Potential energy = mg(2R-r) .........................(2)

and kinetic energy $=\frac{1}{2}m{v_{cm}}^2+\frac{1}{2}{I}_{cm}{\omega }^2....................\left(3\right)$

Moment of inertia of the ball is $I=\frac{2}{5}m{r}^2$and angular speed is $\omega =v/r$

Substitute this in equation (3) we get

$$\begin{gathered} K.E=\frac{1}{2}m{v_{cm}}^2+\frac{1}{2}\left(\frac{2}{5}m{r}^2\right)(v_{cm}/r{)}^2 \\ K.E=\frac{7}{10}m{v_{cm}}^2 \end{gathered}$$

Add equation(2) and (4) to get total energy at the top point of the track

$E_{total}=\frac{7}{10}m{v_{cm}}^2+mg(2R-r)................................\left(5\right)$

Lets consider the ball at the top point of track. There radial force must be equal to weight of the ball.

$$\begin{gathered} \frac{m{v}_{cm}^2}{(R-r)}=mg \\ {v}_{cm}^2=g(R-r)\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \left(6\right) \end{gathered}$$

Substitute the equation (6) in equation (5) we get

$E_{\text{Total}}=\frac{7}{10}mg(R-r)+mg(2R-r)$

From the law of conservation of energy

$mg(h+r)=\frac{7}{10}mg(R-r)+mg(2R-r)$

Cancel mg and solve it.

$$\begin{gathered} h=\frac{7}{10}(R-r)+2(R-r) \\ h=(R-r)\left(2+\frac{7}{10}\right) \\ h=2.7(R-r) \end{gathered}$$